where the limit is taken over partitions $t_0 = 0 \le t_1 \le \cdots \le t_r = T$ whose largest interval is decreasing to $0$ in length. The result belongs to the space $W$.

Hence, to make the above iterated integral fit into the definition (and give a result in $V^{\otimes j}$) it seems to me that one must consider each element $w_1 \otimes \cdots \otimes w_i \in V^{\otimes i}$ as an element of $L(V, V^{\otimes i+1})$ via the map $v \mapsto w_1 \otimes \cdots \otimes w_i \otimes v$. However after doing this I still have trouble establishing the identity.

Because the operation of projecting to a symmetric part is a linear operation from $V^{\otimes k}$ to itself, you can take it inside of the integral. Let's call this symmetric part $dX_{u_1} \cdots dX_{u_k}$. I'll use the symbol $u \cdot v$ to denote the symmetric product of of $u, v \in V$ -- that is, $u \cdot u \cdot u = u \otimes u \otimes u$, and the general product is defined by the polarization identity. For example $u \cdot v = ( u \otimes v + v \otimes u) / 2$. The resulting multiplication is commutative.

Observe also that the region $0 \leq u_1 \leq \ldots \leq u_k \leq t$ is a fundamental domain for the action of $S_k$ on the cube $[0,t]$ -- that is, you can get the whole cube by permuting the variables $u_i$, and none of these regions overlap. Therefore, since there are $k!$ such permutations, we sum over permutations to conclude

But, by Fubini and the multilinearity of the symmetric product, the integral on the right is just $(X_t - X_0)^k$.

I should remark that this same proof in an even simpler setting also produces the $\frac{1}{k!}$ that appears when you prove the Taylor expansion through iterative use of the fundamental theorem of calculus.